If \( f(x)=\ \sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right) \), then
Match List-I with List-II
| List-I | List-II |
| Value of \( x \) | \( \displaystyle\frac{1}{2}f(x) \) |
| (A). \( x=0 \) | (I). \( \tan^{-1}\sqrt{2} \) |
| (B). \( x=\displaystyle\frac{1}{2} \) | (II). \( \tan^{-1}4 \) |
| (C). \( x=1 \) | (III). \( \displaystyle\frac{\pi}{4} \) |
| (D). \( x=2 \) | (IV). \( \tan^{-1}2 \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
Note: The given answer is as per the official NTA Answer Key. However, the answer given by NTA appears to be incorrect. Only (A) - (III) is a correct match, and the correct matching for (B), (C) and (D) is not among the options, as explained below.
Let \( t = 2^x \). Then \( \displaystyle\frac{2^{x+1}}{1+4^x} = \displaystyle\frac{2t}{1+t^2} \), and the standard result is
\( \sin^{-1}\displaystyle\frac{2t}{1+t^2} = 2\tan^{-1}t \) only when \( |t| \le 1 \)
\( \sin^{-1}\displaystyle\frac{2t}{1+t^2} = \pi - 2\tan^{-1}t \) when \( t > 1 \)
The value of \( \sin^{-1} \) is never more than \( \displaystyle\frac{\pi}{2} \), so \( \displaystyle\frac{1}{2}f(x) \) is never more than \( \displaystyle\frac{\pi}{4} \).
(A) x = 0, t = 1: \( \frac12 f(0) = \tan^{-1}1 = \displaystyle\frac{\pi}{4} \) — Correct. (A) matches (III).
(B) \( x=\frac12,\ t=\sqrt2 \): \( \frac12 f = \displaystyle\frac{\pi}{2} - \tan^{-1}\sqrt2 = \tan^{-1}\displaystyle\frac{1}{\sqrt2} \) — Not in List-II. (I) \( \tan^{-1}\sqrt2 \) is more than \( \frac{\pi}{4} \), so it cannot be a value of \( \frac12 f(x) \).
(C) x = 1, t = 2: \( \frac12 f = \displaystyle\frac{\pi}{2} - \tan^{-1}2 = \tan^{-1}\displaystyle\frac12 \) — Not in List-II. (IV) \( \tan^{-1}2 \) is more than \( \frac{\pi}{4} \).
(D) x = 2, t = 4: \( \frac12 f = \displaystyle\frac{\pi}{2} - \tan^{-1}4 = \tan^{-1}\displaystyle\frac14 \) — Not in List-II. (II) \( \tan^{-1}4 \) is more than \( \frac{\pi}{4} \).
As a check at x = 1: \( \displaystyle\frac{2^{2}}{1+4} = 0.8 \), and \( \sin^{-1}0.8 \approx 0.927 \), so \( \frac12 f(1) \approx 0.464 = \tan^{-1}\frac12 \), while \( \tan^{-1}2 \approx 1.107 \).
Advice for the exam: In the real exam, mark the answer that is mathematically correct. If NTA’s provisional answer key marks a different option, challenge it during the answer-key challenge window, giving proper reasoning. NTA does revise its key when a challenge is well justified, so a sound challenge can raise your marks.